The proof follows the one in the Euclidean case in [1]}.
Let \(0\le t_1<t_2\le T\)  as in the hypothesis, let \(N\in \mathbb {N}\)  and set \(\tau :=\frac{t_2-t_1}{N}\)  and \(t^k:=t_1+\tau k\)  for \(k=0,1,\ldots ,N\) .
In the whole interval \([0,T]\)  there is a finite number of vortices. Moreover
in the interval \([t_1,t_2]\) 
the vortices \(a_j\)  (\(j=1,\ldots ,n\) ) remain distinct. Therefore,
taking \(N\)  large enough,
for any interval \([t^{k},t^{k+1})\)  we can find \(n\)  balls \(B_j\)  (\(j=1,\ldots ,n\) ) such that each of them contains exactly one vortice \(a_j(t)\)  for \(t\in [t^{k},t^{k+1})\) .
Now, for any \(k=1,\ldots , N\)  and for any \(j=1,\ldots , d\) , we let \(\psi \in C^1_{c}(B_j)\)  with \(\vert \nabla \psi \vert \le 1\) .
Then, (REF ) with \(X=\nabla ^\perp \psi \) , \(Y = \frac{\partial }{\partial t}\) , together with (REF ) (with \(\omega =V\) ) give
\(\begin{split}&\liminf _{\varepsilon \rightarrow 0}\left(\frac{1}{\vert \log \varepsilon \vert ^2}\int _{B_{j}\times [t^{k},t^{k+1}]}\vert D_{X}u_\varepsilon \vert _g^2 \, \mathrm {vol}_g\, \mathrm {d}t\int _{B_{j}\times [t^k,t^{k+1}]}\vert \partial _t u_\varepsilon \vert ^2_{g} \, \mathrm {vol}_g\, \mathrm {d}t\right) \\&\hspace{113.81102pt}\ge \left| \frac{1}{2}\int _{B_{j}\times [t^k,t^{k+1}]} V\wedge \mathrm {d}\psi  \right|^2\end{split}\) 
Moreover, (REF ) gives that
\(\int _{B_{j}\times [t^{k},t^{k+1}]} V\wedge \mathrm {d}\psi = 2\pi d_j \left( \psi (a_j(t^{k+1}) -a_j(t^k)\right).\) 
Therefore,
\(\begin{split}&\liminf _{\varepsilon \rightarrow 0}\left(\frac{1}{\vert \log \varepsilon \vert ^2}\int _{B_{j}\times [t^k,t^{k+1}]}\vert D_{X}u_\varepsilon \vert _g^2 \mathrm {vol}_g\mathrm {d}t\int _{B_{j}\times [t^k,t^{k+1}]}\vert \partial _t u_\varepsilon \vert ^2_{g}\mathrm {vol}_g\mathrm {d}t\right)\\&\hspace{113.81102pt} \ge \pi ^2\left| \psi (a_j(t^{k+1})) -\psi _\delta (a_j(t^k)) \right|^2.\end{split}\) 
Now, we apply Lemma REF  (specifically () with \(X=Y=\nabla ^\perp \psi _\delta \) )
with \(v_\varepsilon = u_\varepsilon (t)\)  (with \(t\)  fixed in \([t^k,t^{k+1}]\) ) and we get
\(\lim _{\varepsilon \rightarrow 0}\frac{1}{\vert \log \varepsilon \vert }\int _{B_j}\vert D_X u_\varepsilon (t)\vert ^2_g\mathrm {vol}_g= \pi \vert \nabla ^\perp \psi (a_j(t))\vert ^2_g=\pi \vert \nabla \psi (a_j(t))\vert ^2_g,\) 
therefore, since \(\vert \nabla \psi \vert \le 1\) ,
\(&\liminf _{\varepsilon \rightarrow 0}\frac{\tau \pi }{\vert \log \varepsilon \vert }\int _{B_{j}\times [t^k,t^{k+1}]}\vert \partial _t u_\varepsilon \vert ^2_{g}\mathrm {vol}_g\mathrm {d}t\ge \\&\liminf _{\varepsilon \rightarrow 0}\Big (\frac{1}{\vert \log \varepsilon \vert ^2}\int _{B_{j}\times [t^k,t^{k+1}]}\vert D_{X}u_\varepsilon \vert _g^2 \mathrm {vol}_g\mathrm {d}t\int _{B_{j}\times [t^k,t^{k+1}]}\vert \partial _t u_\varepsilon \vert ^2_{g}\mathrm {vol}_g\mathrm {d}t\Big ) \\&\ge \pi ^2\left| \psi (a_j(t^{k+1})) -\psi (a_j(t^k)) \right|^2.\) 

Thus, if we take the supremum over \(\psi \in C^1_{c}(M)\)  with \(\left\Vert  \nabla \psi  \right\Vert _{L^\infty }\le 1\) ,
we get
\(\liminf _{\varepsilon \rightarrow 0}\frac{1}{\vert \log \varepsilon \vert }\int _{B_{j}\times [t^k,t^{k+1}]}\vert \partial _t u_\varepsilon \vert ^2_{g}\mathrm {vol}_g\mathrm {d}t \ge \pi \frac{\operatorname{dist}_g^2(a_j(t^{k+1}),a_j(t^k))}{\tau },\) 
and therefore, summing on \(j=1,\ldots , n\) ,
\(\liminf _{\varepsilon \rightarrow 0}\frac{1}{\vert \log \varepsilon \vert }\int _{\Omega \times [t^k,t^{k+1}]}\vert \partial _t u_\varepsilon \vert ^2_{g}\mathrm {vol}_g\mathrm {d}t \ge \pi \sum _{j=1}^{n}\frac{\operatorname{dist}_g^2(a_j(t^{k+1}),a_j(t^k))}{\tau }.\) 
Given \({\bf a} = (a_1, \ldots ,a_n)\) ,
we define its piecewise constant interpolant \(\bar{\mathbf {a}}_{\tau }\)  by
\(\bar{{\bf a}}_{\tau }(t) := {\bf a}(t^{k+1})\,\,\,\,\,\,\,\hbox{ for }t \in [t^{k},t^{k+1}).\) 
Note that since \({\bf a}\)  is continuous we have that \(\bar{{\bf a}}_\tau \xrightarrow{}a\)  uniformly in \([t_1,t_2]\) .
Moreover, we define
\(\left| \bar{{\bf a}}_{\tau }^{\prime } \right|_g(t):= \left(\sum _{j=1}^n\frac{\operatorname{dist}_g^2(a_j(t^{k+1}),a_j(t^{k}))}{\tau ^2}\right)^{1/2},\,\,\,\,\hbox{ for }t\in [t^{k},t^{k+1}).\) 
Note that, as observed in [2]}, the notation is justified since when one reduces to a linear framework, e.g. \(M=\mathbb {R}^{2n}\) , then \(\left| \bar{\mathbf {a}}_{\tau }^{\prime } \right|_g(t)\)  is the norm of the derivative of the piecewise interpolation of the values \(\mathbf {a}(t^k)\) , for \(k=1,\ldots , N\) .
We set
\(L:= \liminf _{\varepsilon \rightarrow 0}\frac{1}{\pi \vert \log \varepsilon \vert }\int _{\Omega \times [t_1,t_2]}\vert \partial _t u_\varepsilon \vert ^2_{g}\mathrm {vol}_g\mathrm {d}t,\) 
and thus (REF ) rewrites as
\(\int _{t_1}^{t_2}\vert \bar{{\bf a}}_{\tau }^{\prime }\vert ^2(t)\mathrm {d}t \le L.\) 
We assume that \(L<+\infty \)  (otherwise there is nothing to prove). Therefore, there exists \(A\in L^2(t_1,t_2)\)  and a subsequence of \(\tau \)  (not relabelled) such that
\(\vert \bar{{\bf a}}_{\tau }^{\prime }\vert \xrightarrow{}A\,\,\,\,\,\hbox{ weakly in } L^2(t_1,t_2),\) 
and thus
\(\int _{t_1}^{t_2}\vert A\vert ^2 \mathrm {d}t \le L.\) 
Now, we show that \({\bf a}\in H^1(t_1,t_2;M)\)  with metric derivative \(\vert a^{\prime }\vert _g(t)\)  such that
\(\vert {\bf a}^{\prime }\vert _g(t) \le A(t)\,\,\,\,\hbox{ for a.a. }t\in (t_1,t_2).\) 
We fix \(t_1 \le s<t\le t_2\)  and we set
\(\overline{s}_\tau := \max \left\lbrace t^k\colon t^k\le s\right\rbrace \,\,\,\,\,\,\hbox{ and }\,\,\,\,\,\,\,\underline{t}_\tau := \min \left\lbrace t^k\colon t^k\ge t\right\rbrace .\) 
Note that \(\overline{s}_\tau \xrightarrow{}s\)  and \(\underline{t}_\tau \xrightarrow{}t\) .
Then, the triangle inequality gives
\(\sum _{j=1}^n \operatorname{dist}_g(a_j(\overline{s}_\tau ),a_j(\underline{t}_\tau ))\le \int _{\overline{s}_\tau }^{\underline{t}_\tau }\vert \bar{\bf a}^{\prime }\vert _g(r) \mathrm {d}r.\) 
Therefore, if we let \(\tau \searrow 0\) , we get
\(\sum _{j=1}^n \operatorname{dist}_g(a_j(s),a_j(t)) \le \int _{s}^t A(r)\mathrm {d}r,\) 
and thus since \(A\in L^2(t_1,t_2)\) , thanks to Proposition REF  we get that \({\bf a}\in H^1(t_1,t_2;M)\) 
and
\(\vert {\bf a}^{\prime }\vert _g(t) \le A(t)\,\,\,\,\,\hbox{ for a.a. } t\in (t_1,t_2).\) 
Therefore, we conclude
\(\pi \int _{t_1}^{t_2}\vert {\bf a}^{\prime }\vert _g^2(t)\mathrm {d}t \le \liminf _{\varepsilon \rightarrow 0}\frac{1}{\vert \log \varepsilon \vert }\int _{\Omega \times [t_1,t_2]}\vert \partial _t u_\varepsilon \vert ^2_{g}\mathrm {vol}_g\mathrm {d}t. \) 
